Flabbergasting!

$\displaystyle\int_{0}^{\infty}\dfrac{\sin(x^{2})x}{e^{x^{2}}-1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{\sin{x}}{e^x-1}dx =$

$\displaystyle\int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{x^{2n-1}(-1)^{n+1}}{(2n-1)!(e^x-1)}=\quad\text{by expanding the power series for sin}$

$\displaystyle\sum_{n=1}^{\infty}\frac{\zeta(2n)\Gamma(2n)(-1)^{n+1}}{(2n-1)!}dx = \sum_{n=1}^{\infty}(-1)^{n+1}\zeta(2n)\quad\text{by the relation between zeta and gamma}$

$\displaystyle\sum_{n=0}^{\infty}\zeta(2n)x^{2n} = \frac{-x\pi}{2}\cot{(x\pi)}$

Using x = i and a little adjustment for the power of -1 and $\displaystyle\zeta(0) = \frac{-1}{2}$ :

$\displaystyle\sum_{n=1}^{\infty}\zeta(2n)(-1)^{n+1} = \frac{i\pi}{2}\cot{(i\pi)} - \frac{1}{2} = \frac{1}{2}(\pi \coth{(\pi)} -1)$

Remembering the 1/2 from the u sub, $\displaystyle\quad f(x) = x\coth{x}-1$

$I=\int_0^\infty \frac{x\sin x^2}{e^{x^2}-1}dx\\=1/2\int_{0}^{\infty}\frac{\sin u}{e^u-1}du\quad (\mbox{Taking}\ x^2=u)\\ =1/2\sum_{k\ge 1}\int_{0}^{\infty}e^{-ku}\sin u du\\=\frac{1}{2}\sum_{k\ge 1}\frac{1}{k^2+1}$ Now, we use the infinite product formula for $\sin hx$ to get $\sin h\pi x=\pi x\prod_{k\ge 1}\left(1+\frac{x^2}{k^2}\right)\\\implies \ln \sin h\pi x=\ln(\pi x)+\sum_{k\ge 1}\ln\left(1+\frac{x^2}{k^2}\right)$ Differentiating with respect to $x$ implies $\pi \cot h \pi x=\frac{1}{x}+2x\sum_{k\ge 1}\frac{1}{k^2+x^2}\\\implies \frac{1}{2}\sum_{k\ge 1}\frac{1}{k^2+1}=\frac{1}{4}\left(\pi \cot h \pi -1\right)$ which results in the answer $4+1+1-1=\boxed{5}$ .

The Function $f$ is: $f(x) = x\coth(x) -1$ I'm wondering no one has presented a solution even though it has 7 solvers.I invented this and a similar integral and put it up on MSE. Although I solved this one on my own, but the other one is still unsolved and can be found here .( My mse username is kunal)
To begin with the solution, let's consider the following integral: $\large{ I(a) =\int_{0}^{\infty}\dfrac{\sin(x^2)x^a}{e^{x^2}-1}dx }$ Now, it can be expressed as: $\large{I(a) = \int_{0}^{\infty}\sum_{r=1}^{\infty}\sin(x^2)x^{a}e^{-rx^2}dx}$ Also, $\sin(x^2) = \text{Im}[e^{ix^{2}}]$ . Hence, $\large{I(a) =\text{Im}\left[\sum_{r=1}^{\infty}\int_{0}^{\infty}x^{a}e^{-x^{2}(r-i)}dx \right]}$ Note that the summation and the integral can be interchanged because of Fubini's Theorem ( It can be proved using Dominated Convergence Theorem).
Substituting $x^{2}(r-i) = t$ $\large{ I(a) = \text{Im}\left[\sum_{r=0}^{\infty}\int_{0}^{\infty}\dfrac{t^{\frac{a-1}{2}}e^{-t}}{2(r-i)^{\frac{a+1}{2}}}dx \right]}$ BY the definition of Gamma Function, we can write it as $\large{I(a) = \text{Im}\left[\dfrac{\Gamma \left(\dfrac{a+1}{2}\right)}{2}\sum_{r=1}^{\infty}\dfrac{1}{(r-i)^{\frac{a+1}{2}}}\right] }$ Writing $\large{ (r-i) = \sqrt{r^{2}+1}e^{-i\tan^{-1}\left(\frac{1}{r}\right)}}$ Our integral now becomes, $\large{ I(a) = \dfrac{\Gamma\left(\dfrac{a+1}{2}\right)}{2}\text{Im}\left[ \sum_{r=1}^{\infty}\dfrac{e^{i\frac{a+1}{2}\tan^{-1}\left(\frac{1}{r}\right)}}{(r^{2}+1)^{\frac{a+1}{4}}} \right] }$ Taking the Imaginary part, we get $\large{I(a) = \dfrac{\Gamma\left(\dfrac{a+1}{2}\right)}{2}\left[ \sum_{r=1}^{\infty}\dfrac{\sin\left(\dfrac{a+1}{2}\tan^{-1}\left(\dfrac{1}{r}\right)\right)}{(r^{2}+1)^{\frac{a+1}{4}}} \right] }$ Plug $a=1$ to get: $\large{ I(1) = \dfrac{\Gamma(1)}{2}\left[ \sum_{r=1}^{\infty}\dfrac{1}{r^{2}+1}\right] }$ The sum can be evaluated by various methods, including but not limited to Fourier Series, Digamma Functions etc. and comes out to be $\large{ \sum_{r=1}^{\infty}\dfrac{1}{r^{2}+1} = \dfrac{1}{2}(\pi\coth(\pi)-1) }$ Making our required integral to be $\large{ \int_{0}^{\infty}\dfrac{\sin(x^{2})x}{e^{x^{2}}-1}dx =\dfrac{1}{4}(\pi\coth(\pi)-1) \approx 0.538337023734 }$ verified numerically, which completes the solution.